G A 1.8-m3 rigid tank contains steam at 220°C. One-third of the volume is in the liquid phase and the rest is in the vapor form. Determine (a) the pressure of the steam, (b) the quality of the saturated mixture, and (c) the density of the mixture.

Explanation:

tank volume = 1.8 , T = 220°C, = 1/3, = 2/3,

a) The tank contain liquid-vapor mixture, therefore the pressure of the steam in the tank is gotten by the saturated pressure at 220°C

From Table A-4 'saturated water temperature table' we have:

P = 2319.6 KPa ≈ 2320 KPa

b) x = = 2/3 ÷ (1/3 + 2/3)

x = 0.67 = 67%

c) Dmix = Dv * xv + Dl * xl

Dv = 0.6 kg/m3, Dl = 1000 kg/m3

Dmix = 0.6(2/3) + 1000(1/3)

Dmix = 0.4 + 333

Dmix = 333.4 kg/m3

Dmix = 333 kg/m3

answer:

wth

explanation: